Eigenvalues of Linear operator

**charikaar** · May 10th 2010, 11:29 AM

Hello,
Find all the eigenvalues and determine with a reason whether L is self-adjoint.
$d/dt<img src=$ " alt="d/dt

" />----------> $L_2[0,1]$ where the domain D is given by D{f belongs to $L_2[0,1]$ : f is continuously differentiable and f(0)=0)}
I get $f(x)=e^{\lambda x}$ after solving first order linear differential equation. How do i find $\lambda$ ?
thanks

**HallsofIvy** · May 11th 2010, 05:00 AM

Originally Posted by charikaar

Hello,
Find all the eigenvalues and determine with a reason whether L is self-adjoint.
$d/dt<img src=$ " alt="d/dt

" />----------> $L_2[0,1]$ where the domain D is given by D{f belongs to $L_2[0,1]$ : f is continuously differentiable and f(0)=0)}
I get $f(x)=e^{\lambda x}$ after solving first order linear differential equation. How do i find $\lambda$ ?
thanks

our operator is just differentiation so, yes, $\frac{d e^{\lambda x}}{dx}= \lambda e^{\lambda x}$ . More generally, if $\frac{dy}{dx}= \lambda y$ , then $y= Ce^{\lambda 0}$ $e^{\lambda x}$ is an "eigenvector" and $\lambda$ is an eigenvalue. As for finding $\lambda$ , you don't! That equation is true for all $\lambda$ - any real number is an eigenvalue.

As for whether or not it is "self adjoint", the definition of self-adjoint is that <Af, g>= <f, Ag> for any vectors f, g. Here, I presume you are using the "standard" inner product, $<f, g>= \int_0^1 f(x)g(x)dx$ so that we want $\int_0^x \frac{df}{dx} g dt$ . Use "integration by parts" with u= g and dv= (df/dx)dx. Then du= (dg/dx)dx and v= f.

$\int u dv= uv- \int vdu= f(x)g(x)- \int f (dg/dx) dx= <f, dg/dx>$ provided f(1)g(1)- f(0)g(0)= 0.

**Danny** · May 11th 2010, 05:46 AM

I'm curious on one thing. Don't we require that $f(0) = 0$ ?

**charikaar** · May 11th 2010, 07:56 AM

Thanks for your help.

$L_2[0,1]$ is the Hilbert space of complex square integrable functions on [0,1] and D={f belongs to $L_{2}[0,1]$ : f is continuously differentiable and f(0)=1)}. so $\lambda$ includes complex numbers also? L is not self-adjoint because there is minus in front of the integral right?

f(0)=1 not zero.

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